Exercise 12D
Question 1
Find the curved surface area and whole surface area for the cylinders shown below. Measures are in centimetres.
(a)
h=5 ,r=7, $\pi =\frac{22}{7}$
∴C.S.A=2πrh
$=2\times \frac{22}{7}\times 7\times 5 $
=220cm2
T.S.A=2πrh+(πr2×2)
$=220+\frac{22}{7}\times 7\times 7\times 2$
=220+(154×2)
=220+308=528cm2
(b)
h=8 ,r=5 ,π=3.14
∴C.S.A=2πrh
=2×3.14×5×8
=251.20cm2
T.S.A=2πrh+(πr2×2)
=251.20+{(3.14×52)×2}
=251.20+157
=408.20cm2
(c)
h=10 ,r=2, π=3.14
C.S.A=2πrh
=2×3.14×2×10
=125.60cm2
T.S.A=2πrh+2πr2
=125.6+(2×3.14×22)
=125.6+(2×3.14×4)
=125.6+25.12=150.72cm2
(d)
h=3 ,r=7, $\pi=\frac{22}{7}$
C.S.A=2πrh
$=2\times \frac{22}{7}\times 7\times 3$
=132cm2
T.S.A=2πrh+2πr2
$=132+\left(2\times \frac{22}{7}\times 7\times 7\right)$
=132+308=440cm2
Question 2
The table given below contains information about a few cylinders. Complete it.
| Radius | Height | Curved Surface Area | Area of the base | Whole Surface | |
|---|---|---|---|---|---|
| (a) | 7 cm | 12 cm | |||
| (b) | 2.8 m | 5 m | |||
| (c) | 35 mm | 50 mm |
Sol :
(a)
r=7cm , h=12cm
T.S.A=2πrh+2πr2
$=\left(2\times \frac{22}{7}\times 7\times 12\right)+\left(2\times \frac{22}{7}\times 7 \times 7\right)$
=528+308=836cm2
Area of the base=πr2
$=\frac{22}{7}\times 7\times 7$
=154cm2
C.S.A=2πrh
$=2\times \frac{22}{7} \times 7\times 12$
=528cm2
(b)
r=2.8m ,h=5m
C.S.A=2πrh
=2×2.8×5×3.14
=87.92m2
T.S.A=2πrh+2πr2
=87.92+2×3.14×2.8×2.8
=87.92+49.22
=137.15m2
Area of the base=πr2
=3.14×2.8×2.8
=24.61m2
(c)
r=35mm ,h=50mm
C.S.A=2πrh
$=2\times \frac{22}{7}\times 35 \times 50$
=11000m2
Area of the base=πr2
$=\frac{22}{7}\times 35\times 35$
=3850m2
T.S.A=2πrh+2πr2
=11000+(2×3850)
=18700m2
Question 3
Find the missing dimension for a cylinder.
(a) Radius: Curved surface area $=220 \mathrm{~cm}^{2}$. Height =10 cm
Sol :
(a)
C.S.A=220cm2 ,h=10cm , r=?
⇒C.S.A=2πrh
$r=\frac{\text{C.S.A}}{2 \pi h}=\frac{220}{2\times \frac{22}{7}\times 10}$
$r=\frac{220\times 7}{2\times 22\times 10}=\frac{7}{2}=3.5$cm
(b) Height: Curved surface area = $3168 \mathrm{~cm}^{2}$, Radius = 21 cm
Sol :
(b)
C.S.A=2πrh
$h=\frac{\text{C.S.A}}{2\pi r}=\frac{3168}{2\times \frac{22}{7}}\times 21$
$=\frac{3168}{2\times 22\times 3}$
h=24cm
(c) Height: Radius=5 cm. Total surface area $=120 \pi\mathrm{cm}^{2}$
Sol :
(c)
T.S.A=120π , r=5 , h=?
T.S.A=2πrh+2πr2
120π=2π(rh+r2)
120π=2π(5h+52)
60=5h+25
5h=60-25=35
h=7cm
Question 4
The circumference of a right circular cylinder is 440 cm and its height is 5 cm. Find the lateral surface area of the cylinder.
Sol :
c=440cm , h=5cm
ATQ ,
2πr=440
$r=\frac{440}{2\times \pi}$
$r=\frac{440}{2\times \frac{22}{7}}$
$=\frac{440\times 7}{2\times 22}$
=70cm
∴L.S.A=2πrh
$=2\times \frac{22}{7}\times 70\times 5$
=2200cm2
Question 5
A cylindrical soup can is 15.4 cm in diameter and 20 cm in length. What is the area of the label that covers the side of the can?
Sol :
d=15.4cm
∴$r=\frac{15.4}{2}$cm ,h=20cm
⇒Lateral Surface Area=2πrh
$=2\times \frac{22}{7}\times \frac{15.4}{2}\times 20$
$=2\times \frac{22}{7}\times \frac{154}{2\times 10}\times 20$
=968cm2
Question 6
In a building, there are 30 cylindrical pillars. The radius of each pillar is 42 m and height is 7 m. Find the total cost of painting the curved surface area of all pillars at the rate of ₹ 10 per m2?
Sol :
⇒C.S.A of a pillar=2πrh
$=2\times \frac{22}{7}\times 42\times 7$
=1848m2
C.S.A of 30 pillar=1848×30
=55440m2
∴Paint cost for 30 pillar's=55440×10
=554400
Question 7
A cylinder has a diameter 20 cm and height 7 cm. Explain whether tripling the height would have the same effect on the surface area as tripling the radius or not.
Sol :
∴T.S.A for 1st case=2πr(r+h1)
=2×3.14×10×(10+21)
=1946.8 cm2
∴T.S.A to 2nd case=2πr1(r1+h)h
=2×3.14×30×(30+7)
=6970.8 cm2
∴Both case surfaces are will be different.
Question 8
The diameter of a garden roller is 1.8 m and it is 2.8 m long. How much area will it cover is 25 revolutions in levelling a field?
Sol :
d=1.8m , h=2.8m , $r=\frac{1.8}{2}=0.9$m
⇒C.S.A=2πrh
$=2\times \frac{22}{7}\times 0.9\times 2.8$
$=2\times \frac{22}{7}\times \frac{9}{10}\times \frac{28}{10}$
$=\frac{2\times 22\times 9\times 4}{100}=\frac{396}{25}$m2
∴1 revolution need$=\frac{396}{25}$m2 area
∴25 revolution need$=\frac{396}{25}\times 25$=396m2
Question 9
The diameter of a road roller is 90 cm and is 2 m 10 cm in length. If it takes 400 revolutions to level a playground , find the cost of levelling the ground at 2.25 per square meter.
Sol :
⇒1 revolution=C.S.A of roller=2πrh
$=2\times \frac{22}{7}\times 45 \times 210$
=59400 cm2 or 5.94 m2
If takes 400 revolution to level 5.94 m2 area
It takes 400 revolutions , so total area of ground = 5.94×400=2376 m2
∴Cost of levelling at 2.25 for 1m2
∴Cost of levelling
(2376×2.25)₹=(2376×1) m2
5346 ₹=2376 m2
Question 10
The inner diameter of a 15 m deep circular well is 4.2 m. Find the cost of plastering its inner covered surface at the rate of 40 per m2
Sol :
Cost of plastering of 1m2 area=40
∴Cost of plastering of 198 m2 area=198×40=7920
Question 11
The inner circumference of a hollow cylindrical pipe is 4.5 m and outer circumference is 4.6 m. Find the cost of painting both its inner and outer sides at the rate of 20 per m2, if it is 8 m tall
Sol :
∴$r=\frac{4.5}{2\pi}$
$=\frac{4.5}{2\times 3.14}=\frac{4.5}{6.28}$
=0.715m
Outer circumference=2πR=4.6
$R=\frac{4.6}{2\pi}=\frac{4.6}{2\times 3.14}$
=0.731m
Inner C.S.A=2πrh
$=2\times \frac{22}{7}\times 0.731\times 8$
=35.95m2
Outer C.S.A=2πRh
$=2\times \frac{22}{7}\times 0.731\times 8$
=36.75m2
∴Total C.S.A=35.95+36.75=72.7m2
∵Cost of painting 1m2=20
∴Cost of painting hollow cylinder=72.7×20=1454
Question 12
The external diameter of a 20 cm long and 1 cm thick hollow iron pipe is 25 cm. Determine the whole surface of the iron pipe.
Sol :
∴Area of one circular end=2(πR2-πr2)
$=2\times \frac{22}{7} \left\{(12.5+11.5)(12.5-11.5)\right\}$
$=\frac{44}{7}\times 24$ cm2
∴T.S.A of Iron pipe$=\left(\frac{44\times 480}{7}\right)+\left(\frac{44}{7}\times 24\right)$
$=\frac{44}{7}\left(480+24\right)$
$=\frac{44}{7}\times 504$
=44×72=3168 cm2
No comments:
Post a Comment